Question

高一数学题2道 50分！要过程

1，已只函数f(x)=2cosx(sinx-cosx)+1,x属于R (1)求函数f(x)的最小正周期。（2）求函数f(x)在区间[π/8,3/4π]上最小值和最大值

2，已知sinx+siny=1/3,(1)求sinx的取值范围。（2）求a=siny-cos的平方x的最值

Answer 1

f(x)=2cosx(sinx-cosx)+1
＝2sinxcos-2cosx^2+1
=sin2x-cos2x
=根号2*sin(2x-π/4)
T=2π/2=π
x∈[π/8,3π/4]，2x-π/4∈[0，5π/4],
sin(2x-π/4)∈[-根号2/2，1】,
最小值为－1，最大值为根号2。

2. -1≤siny=1/3-sinx≤1,且-1≤sinx≤1
解得：-2/3≤sinx≤1
a=siny-(cosx)^2=1/3-sinx+(sinx)^2-1
=(sinx)^2-sinx-2/3
=(sinx-1/2)^2-11/12
当sinx=1/2时，最小值是-11/12
当sinx=-2/3时，最大值是4/9

Answer 2

解：1.
f(x)=2osx(sinx-cosx)+1=2sinx*cosx-2(cosx)^2+1=sin2x-cos2x
=根号2(sin2xcosπ/4-cos2xsinπ/4)=根号2sin(2x-π/4)
那么函数的最小正周期为π
在[π/8,3/4π]中，π/8<=x<=3/4π
π/4<=2x<=3/2π
0<=2x-π/4<=5/4π
那么在该区间最小值是0，最大值是根号2

2.由于sinx+siny=1/3,sinx=1/3-siny,由于siny的取值是[0,1],则
(1/3)-1<=sinx<=(1/3)+1,-2/3<=sinx<=4/3,由于sinx<=1,则
-2/3<=sinx<=1
由于-2/3<=sinx<=1
4/9<=(sinx)^2<=1
-1<=-(sinx)^2<=-4/9
1-1<=1-(sinx)^2<=1-4/9
0<=(cosx)^2<=5/9
-5/9<=(cosx)^2<=0
而-1<=siny<=1,那么
-1-5/9<=siny-(cosx)^2<=1+0
-14/9<=siny-(cosx)^2<=1